Question

The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?

Answer 1

Given:

The distance of the star is,$=4.29ly$

We know that one light year is the distance travelled by light in one year.

1 light year = $3\times {10}^8\times 365\times 24\times 60\times 60$

=$94608\times {10}^{11}m$

Therefore, the distance in $4.29ly$ can be written as:

$4.29ly=405868.32\times {10}^{11}m$

We know that $1Parsec$ is also the unit of distance and its value is:

$1parsec=3.08\times {10}^{16}m$

Now, to express the distance in terms of parsec.

$4.29ly=\frac{405868.32\times {10}^{11}}{3.08\times {10}^{16}}=1.32parsec$

The diameter of the Earth's orbit is,

$d=3\times {10}^{11}m$

The angle made by the star on the Earth's orbit can be given by:

$\theta =\frac{3\times {10}^{11}}{405868.32\times {10}^{11}}=7.39\times {10}^{-6}rad$

But, the angle covered in $1sec=4.85\times {10}^{-6}rad$

So, the parallax in viwing the star at two different positions of the Eart's orbit:

for, $7.39\times {10}^{-6}rad=\frac{7.39\times {10}^{-6}}{4.85\times {10}^{-6}}=1.52"$